Answer
(a) NO
(b) $(4,-3)$
(c) $(14,2)$
(d) all real numbers except $x=6$
(e) $(-2,0)$
(f) $\left(0,-\frac{1}{3}\right)$
Work Step by Step
(a) To determine if $(3,14)$ is on the graph of $f$, substitute the $x$ and $y$ coordinates into the function. If the coordinates satisfy the function, then the point is on the graph of $f$.
$f(x)=\dfrac{x+2}{x-6}$
$14=\dfrac{3+2}{3-6}$
$14\ne\dfrac{5}{-3} $
Thus, $(3, 14)$ is not on the graph of $f$.
(b) Substitute $4$ to $x$ to obtain:
$f(x)=\dfrac{x+2}{x-6}$
$f(4)=\dfrac{4+2}{4-6}$
$f(4)=-3$
The point on the graph are is $(4,-3)$
(c) Substitute $2$ to $f(x)$ then solve the equation to obtain:
$f(x)=\dfrac{x+2}{x-6}$
$2=\dfrac{x+2}{x-6}$
$2(x-6)=x+2$
$2x-12=x+2$
$2x-x=2+12$
$x=14$
Therefore, the point on the graph is $(14,2)$
(d) The function is defined for all real numbers except when $x=6$ as it makes the denominator $0$.
Hence, the domain is all real numbers except $x=6$
(e) To find the $x$-intercepts, subsitute $y=0$ to obtain:
$f(x)=\frac{x+2}{x-6}$
$0=\frac{x+2}{x-6}$
$0=x+2$
$-2=x$
Therefore, the $x$ intercept is $(-2,0)$
(f) To find the $y$-intercept, set $x=0$ to obtain:
$f(x)=\frac{x+2}{x-6}$
$y=\frac{0+2}{0-6}$
$y=-\frac{1}{3}$
Therefore, the $y$-intercept is $\left(0,-\frac{1}{3}\right)$
